How do you find the integral ln(x) x^(3/2) dx?

1 Answer
Jun 18, 2015

To do integration by parts:

uv = intvdu

I would pick a u that gives me an easy time differentiating, and a dv that I would easily be able to integrate multiple times if necessary. If the same integral comes back, or variables aren't going away, either it's cyclic or you should switch your u and dv.

Let:
u = lnx
du = 1/xdx
dv = x^(3/2)dx
v = 2/5x^(5/2)

=> 2/5x^(5/2)lnx - int2/5(x^(5/2)/x)dx

Nice! That's not too bad.

= 2/5x^(5/2)lnx - int2/5x^(3/2)dx

= 2/5x^(5/2)lnx - 4/25x^(5/2) + C

= 10/25x^(5/2)lnx - 4/25x^(5/2) + C

= x^(5/2)[10/25lnx - 4/25] + C

= 2/25x^(5/2)[5lnx - 2] + C